Question: Find the slope and y-intercept of the line that is ${\text{parallel}}$ to $\enspace {y = \dfrac{1}{2}x + 3}\enspace$ and passes through the point ${(-8, -6)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Solution: Parallel lines have the same slope. The slope of the blue line is ${\dfrac{1}{2}}$ , so the equation of our parallel line will be of the form $\enspace {y = \dfrac{1}{2}x + b}\enspace$ We can plug our point, $(-8, -6)$ , into this equation to solve for ${b}$ , the y-intercept. $-6 = {\dfrac{1}{2}}(-8) + {b}$ $-6 = -4 + {b}$ $-6 + 4 = {b} = -2$ The equation of the parallel line is $\enspace {y = \dfrac{1}{2}x - 2}\enspace$. ${m = \dfrac{1}{2}, \enspace b = -2}$